. Linear Algebra Final Solutions. (pts) Let A =. (a) Find an orthogonal matrix S a diagonal matrix D such that S AS = D. (b) Find a formula for the entries of A t,wheret is a positive integer. Also find the vector lim A t. t (a) Let f A (λ) =det(a λi )=det λ λ = λ λ +λ += (λ ) (λ +) =.Wehavetheeigenvalues are with multiplicity with multiplicity. For λ =, the eigenspace E = ker = ker = ker = span.
Therefore, forms an orthonormal basis of E. For λ =, the eigenspace E = ker ( ) ( ) ( ) = ker = ker = span,. Therefore, v = v = form a eigenbasis of E. Using Gram-schmidt process on v v to get an orthonormal eigenbasis w w for E,wehave w = v k v k =, with u = v ( w v ) w = = on h, we have k u k = w = v ( w v ) w k v ( w v ) w k = u k u k = =.
Set S = D =. We have S is orthogonal D is diagonal. Therefore, we have D = S AS where S is orthogonal D is diagonal. (b)from(a),wehavea = SDS. Since S is orthogonal, we have S = S T. Hence, for a positive integer t, A t = SD t S = SD t S T t = ( ) t ( ) t t +( ) t t ( ) t t ( ) t = t ( ) t t +( ) t t ( ) t t ( ) t t ( ) t t +( ) t. Therefore, lim t At t +( ) t t ( ) t t ( ) t = lim t t ( ) t t +( ) t t ( ) t t ( ) t t ( ) t t +( ) t = lim ( )t. t ( ) t That means the limit does not exist since you have two different limit points.. (pts) Let A =. (a) Find a singular value decomposition for A. (b) Describe the image of the unit circle under the linear transformation T ( x) =A x.
4 (a) The singular values are the square roots of the eigenvalues of A T A =. Let f A T A (λ) = det A T A λi = µ λ det =(λ ) (λ ) =. We have the λ eigenvalues of A T A are λ = λ =. Therefore, the singular values of A are σ = p λ = σ = p λ =. For σ = µ, the eigenspace E =ker µ ker unit vector =span ½ v = = ¾. Therefore, the nonzero forms an µ orthonormal basis of E.Forσ µ =, the eigenspace ½ E =ker =ker =span Therefore, the nonzero unit vector v = ¾. forms an orthonormal basis of E.Let u = A v = σ = µ
5 u = A v = σ =. µ Since A is a matrix, we know that U will be a matrix. So, we need to exp { u, u } into a orthonormal basis of R. That means we need to find a u such that u, u u form an orthonormal basis of R.Choose w =. Obviously, w is not a linear combination of u u.so, u, u w form a basis of R.Since u u are orthogonal unit vectors already. We use Gram-Schmidt process on w to get u = w ( u w ) u ( u w ) u k w ( u w ) u ( u w ) u k = And, we have u, u u form an orthonormal basis of R. Let V = v v,. U = u u u = Σ = σ σ. Therefore, we get a singular value decomposition(svd) for A = UΣV T.
(b) From (a), we have T ( v ) = A v = = µ T ( v ) = A v = =. µ The unit circle in R consists of all vectors of the form x = c v + c v where c + c =. Therefore,theimageoftheunitcircleunderT consists of the vectors T ( x) =c T ( v )+c T ( v ) where c + c =. That means T ( x) =c + c where c + c =. This is an ellipse with the semimajor axe is the line generated by with the length σ = the semiminor axe is the line generated by. (pts) Let q be a quadratic form q (x,x )=9x 4x x +x. (a) Determine the definiteness of q. with the length σ =.
(b) Sketch the curve defined by q (x,x )=. Draw label the principal axes, label the intercepts of the curve with the principal axes, give the formula of the curve in the coordinate system defined by the principal axes. 7 9 (a) Let A = [Method ] Calculate.Wehaveq ( x) = x A x. A () =det([9])=9> µ 9 A () =det =5>. By the theorem in the textbook, we have q is positive definite. [Method ] Let λ λ be the two eigenvalues of A with associated eigenvectors v v, respectively. We have λ λ =det(a) = 54 4=5> λ + λ =tr(a) =9+=5>. This implies that λ > λ >. Moreover, we have q ( x) =λ c + λ c where x = c v + c v. That means q ( x) > for all x =. By the definition of definiteness, A is positive definite. 9 (b) Let A =. We have q ( x) = x A x. Set = µ 9 λ f A (λ) = det(a λi ) = det = λ λ 5λ +5=(λ ) (λ 5). We get the eigenvalues λ = λ =5.Forλ =, the eigenspace µ 9 E = ker µ = ker 4 =span ½ ¾.
8 ½ ¾ Therefore, 5 forms an orthonormal basis of E. For λ =5, the eigenspace µ 9 5 E 5 = ker 5 µ ½ ¾ 4 = ker =span. ½ ¾ Therefore, 5 forms an orthonormal basis of E 5. Hencewehaveanorthonormaleigenbasis 5 5 v = v =. 5 5 Moreover, the curve can be re-written as c +5c =. Thatmeansthecurveisaellipseinc c coordinates system where the principal axes are E E 5 which are generated by v v, respectively. The graph is
4. (pts) Decide whether the matrix A = is diagonalizable. If possible, find an invertible S a diagonal D such that S AS = D. Using f A (λ) =det(a λi )=det λ λ = λ ( λ) λ =, we have eigenvalues are,. For λ =,the eigenspace E = ker(a I )=ker = ker =span. For λ =, the eigenspace E = ker(a I )=ker = ker =span. Since dim (E )+dim(e )=+=<, we have no eigenbasis of R in this case. This implies that A is not diagonalizable. 5. (pts) Find the trigonometric function of the form f (t) =c + c sin (t)+c cos (t) that best fits the data points (, ), π,, (π, ) π,, using lease squares. We want to find a f (t) =c + c sin (t)+c cos (t) such that f () =, f π =, f (π) = f π =. These conditions give the 9
system of linear equations c + c sin () + c cos () = f () = c + c sin π + c cos π = f π = c + c sin (π) + c cos (π) = f (π) = c + c sin π + c cos, π = f π = or, c + c = c + c =. c c = c c = Let A =, x = c c b = c write the system as A x = b. Since rref (A) = ker (A) ={}. So, the unique least-squares solution of A x = b is x = A T A A T b = = 4. We can,wehave =. Moreover, the trigonometric function, f (t) =+ sin (t) cos (t), best fits the data points (, ), π,, (π, ) π, in the leastsquares sense.
. (pts) Given a matrix A = 7 5. 8 5 (a) Find a basis of kernel of A dim (ker (A)). (b) Find a basis of image of A dim (im (A)). By Gauss-Jordan elimination, we have A = 7 5 8 5 7 7 7 = rref (A). (a) Assume x = x x x x 4 ker A. Then we have A x =. By Gauss-Jordan elimination(which implies ker (A) = ker rref (A)), we have x x x =. x 4 Assume x = s x 4 = t for all s, t R. Wehavesolutions of the system, x s + t x x = s + t s = s + t. x 4 t
Assume v = v = span { v, v }. Moreover, we check that A ( v )= 7 5 8 5 A ( v )= 7 5 8 5.Thatmeansker (A) = = This tells us that { v, v } ker (A) which implies ker (A) = span { v, v }.Tocheck v v are linearly independent, we assume that c v + c v =,or =c + c = c + c c + c c c.. This implies that c = c =. That proves that v v are linearly independent. Now, we can say v, v form a basis of ker (A). And,dim (ker (A)) equals the number of vectors in a basis. So, dim (ker (A)) =. (b) Set v, v, v, v 4 be the column vectors of A. Set w, w, w, w 4 bethecolumnvectorsofrref A. Weknowthat im (A) =span{ v, v, v, v 4 }. From our reduced row-echlon form, we can read two relations of w, w, w, w 4, w = w + w w 4 = w w. This implies we have the same relation of v, v, v, v 4 which are v = v + v
v 4 = v v. Therefore, we can conclude that v, v are linearly independent im (A) =span{ v, v, v, v 4 } =span{ v, v }. Now, we can say that v =, v = 7 form a basis of im (A). And,dim (im (A)) equals the number of vectors in a basis. So, dim (im (A)) =. 7. (pts) Let T from R to R be the reflection in the plane given by the equation x +x +x =. (a) Find the matrix B of this transformation with respect to the basis v =, v =, v =. (b) Use your answer in part (a) to find the stard matrix A of T. (a) Let P be the plane given by the equation x +x +x =. We observe that v v arebothontheplanep.sincet is areflection, it keeps v v unchanged, that is, T ( v )= v T ( v )= v. And, v is the normal vector of P. That means v is perpendicular to the plane P. Therefore, since T is a reflection in P,wehaveT ( v )= v. Withrespect to the basis B = { v, v, v },wehave[t ( v )] B = [T ( v )] B = [T ( v )] B =.So B B B = [T ( v )] B [T ( v )] B [T ( v )] B =. B,
4 (b) Write S = v v v =. Bythetheorem in the textbook, we have A = SBS. To find S,weuse Gauss-Jordan elimination on 4 4 4 8 4 4 4 4 5 4 4 4 4. 4 4 4 Hence, S = 4 7 5 4 4 7 7 7 4 7 4 A = SBS 7 7 7 = 7 7 7. 7 7 7 8. (pts) Consider a linear transformation T from V to W. (a) For f,f,,f n V,ifT(f ),T (f ),,T (f n ) are linearly independent, show that f,f,,f n are linearly independent. (b) Assume that f,f,,f n form a basis of V. If T is an isomorphism, show that T (f ),T (f ),,T (f n ) is a basis of W.
(a) Assume a f + a f + + a n f n =. By applying T on both sides, we have ³ T (a f + a f + + a n f n )=T =. Since T is a linear transformation, a T (f )+ + a n T (f n )=T (a f + + a n f n )=. Since T (f ),T (f ),,T (f n ) are linearly independent, we have a = a = = a n =. That tells us that f,f,,f n are linearly independent. (b) To show that T (f ),T (f ),,T (f n ) are linearly independent, we assume c T (f )+c T (f )+ + c n T (f n ) =. Hence, we have T (c f + + c n f n )=since T is a linear transformation. Since T is an isomorphism from V to W,we have T is invertible, that means, ker (T )={}. This implies c f + + c n f n =. Moreover, f,f,,f n form a basis of V. This condition forces that c = c = = c n =. So, T (f ),T (f ),,T (f n ) are linearly independent. Since T is an isomorphism from V to W,wehaveT is an isomorphism from W to V such that T (T (g)) = g for all g W. For all g W, we have T (g) V. Since f,f,,f n form a basis of V,thereexistt,t,,t n such that T (g) =t f + + t n f n. Therefore, g = T (T (g)) = T (t f + + t n f n )=t T (f )+ t n T (f n ). That means g is a linear combination of T (f ),T (f ),,T (f n ). And, obviously, T (f ),T (f ),,T (f n ) are all in W. Thus, we can conclude that W =span{t (f ),T (f ),,T (f n )}. So, T (f ),T (f ),,T (f n ) form a basis of W 5